Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $a = \dfrac{x + 4}{3x^3 + 24x^2 + 48x} \times \dfrac{x^3 - 4x^2 - 45x}{-3x - 15} $
Explanation: First factor out any common factors. $a = \dfrac{x + 4}{3x(x^2 + 8x + 16)} \times \dfrac{x(x^2 - 4x - 45)}{-3(x + 5)} $ Then factor the quadratic expressions. $a = \dfrac {x + 4} {3x(x + 4)(x + 4)} \times \dfrac {x(x + 5)(x - 9)} {-3(x + 5)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {(x + 4) \times x(x + 5)(x - 9) } { 3x(x + 4)(x + 4) \times -3(x + 5)} $ $a = \dfrac {x(x + 5)(x - 9)(x + 4)} {-9x(x + 4)(x + 4)(x + 5)} $ Notice that $(x + 4)$ and $(x + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {x(x + 5)(x - 9)\cancel{(x + 4)}} {-9x\cancel{(x + 4)}(x + 4)(x + 5)} $ We are dividing by $x + 4$ , so $x + 4 \neq 0$ Therefore, $x \neq -4$ $a = \dfrac {x\cancel{(x + 5)}(x - 9)\cancel{(x + 4)}} {-9x\cancel{(x + 4)}(x + 4)\cancel{(x + 5)}} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $a = \dfrac {x(x - 9)} {-9x(x + 4)} $ $ a = \dfrac{-(x - 9)}{9(x + 4)}; x \neq -4; x \neq -5 $